D.81 Hydrogen ground state Stark effect

This note derives the Stark effect on the hydrogen ground state. Since spin is irrelevant for the Stark effect, it will be ignored.

The unperturbed ground state of hydrogen was derived in chapter 4.3. Following the convention in perturbation theory to append a subscript zero to the unperturbed state, it can be summarized as:

\begin{displaymath}
H_0\psi_{100,0} = E_{100,0}\psi_{100,0}
\qquad H_0 = -\f...
...+V
\qquad \psi_{100,0}=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}
\end{displaymath}

where $H_0$ is the unperturbed hydrogen atom Hamiltonian, $\psi_{100,0}$ the unperturbed ground state wave function, $E_{100,0}$ the unperturbed ground state energy, 13.6 eV, and $a_0$ is the Bohr radius, 0.53 Å.

The Stark perturbation produces a change $\psi_{100,1}$ in this wave function that satisfies, from (A.243),

\begin{displaymath}
(H_0-E_{100,0})\psi_{100,1} = -(H_1-E_{100,1})\psi_{100,0}
\quad H_1 = e{\cal E}_{\rm {ext}}z
\end{displaymath}

The first order energy change $E_{100,1}$ is zero and can be dropped. The solution for $\psi_{100,1}$ will now simply be guessed to be $\psi_{100,0}$ times some spatial function $f$ still to be found:

\begin{displaymath}
\left(H_0-E_{100,0}\right)(f\psi_{100,0}) =
-e{\cal E}_{...
..._{100,0}
\qquad H_0 = -\frac{\hbar^2}{2m_{\rm e}}\nabla^2+V
\end{displaymath}

Differentiating out the Laplacian $\nabla^2$ of the product $f\psi_{100,0}$ into individual terms using Cartesian coordinates, the equation becomes

\begin{displaymath}
f \left(H_0-E_{100,0}\right)\psi_{100,0}
- \frac{\hbar^2...
...abla^2 f)\psi_{100,0}
=
-e{\cal E}_{\rm ext}z\psi_{100,0}
\end{displaymath}

The first term in this equation is zero since $H_0\psi_{100,0}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $E_{100,0}\psi_{100,0}$. Also, now using spherical coordinates, the gradients are, e.g. [40, 20.74, 20.82],

\begin{displaymath}
\nabla f = \frac{\partial f}{\partial r} {\hat\imath}_r + ...
...la \psi_{100,0} = - \psi_{100,0} \frac{1}{a_0} {\hat\imath}_r
\end{displaymath}

Substituting that into the equation, it reduces to

\begin{displaymath}
\frac{\hbar^2}{m_{\rm e}}
\left(\frac{1}{a_0}\frac{\part...
...right)
\psi_{100,0}
=
-e{\cal E}_{\rm ext}z\psi_{100,0}
\end{displaymath}

Now $z$ $\vphantom0\raisebox{1.5pt}{$=$}$ $r\cos\theta$ in polar coordinates, and for the $r$-​derivative of $f$ to produce something that is proportional to $r$, $f$ must be proportional to $r^2$. (The Laplacian in the second term always produces lower powers of $r$ than the $r$-​derivative and can for now be ignored.) So, to balance the right hand side, $f$ should contain a highest power of $r$ equal to:

\begin{displaymath}
f = - \frac{m_{\rm e}e{\cal E}_{\rm ext}a_0}{2\hbar^2} r^2 \cos\theta + \ldots
\end{displaymath}

but then, using [40, 20.83], the $\nabla^2f$ term in the left hand side produces an $e{\cal E}_{\rm {ext}}a_0\cos\theta$ term. So add another term to $f$ for its $r$-​derivative to eliminate it:

\begin{displaymath}
f = - \frac{m_{\rm e}e{\cal E}_{\rm ext}a_0}{2\hbar^2} r^2...
...frac{m_{\rm e}e{\cal E}_{\rm ext}a_0^2}{\hbar^2} r \cos\theta
\end{displaymath}

The Laplacian of $r\cos\theta$ $\vphantom0\raisebox{1.5pt}{$=$}$ $z$ is zero so no further terms need to be added. The change $f\psi_{100,0}$ in wave function is therefore

\begin{displaymath}
\psi_{100,1}= - \frac{m_{\rm e}e{\cal E}_{\rm ext}a_0}{2\h...
...pi a_0^3}}
\left(r^2 + 2 a_0 r\right) e^{-r/a_0}\cos \theta
\end{displaymath}

(This small perturbation becomes larger than the unperturbed wave function far from the atom because of the growing value of $r^2$. It is implicitly assumed that the electric field terminates before a real problem arises. This is related to the possibility of the electron tunneling out of the atom if the potential far from the atom is less than its energy in the atom: if the electron can tunnel out, there is strictly speaking no bound state.)

Now according to (A.243), the second order energy change can be found as

\begin{displaymath}
E_{100,2} = \langle\psi_{100,0}\vert H_1\psi_{100,1}\rangle
\qquad H_1 = e{\cal E}_{\rm {ext}} r \cos \theta
\end{displaymath}

Doing the inner product integration in spherical coordinates produces

\begin{displaymath}
E_{100,2} = - \frac{9m_{\rm e}e^2{\cal E}_{\rm ext}^2a_0^4}{4\hbar^2}
\end{displaymath}